text:
What's the functional difference between PAPI and VASI?
PAPI and VASI seem to be very similar in the the information they provide - i.e. are you on the glide slope or off it. Is there a functional difference between the two? If there is a difference, and one is superior to the other, is there a construction cost savings where an airport might chose one over the other?

A PAPI and a VASI are very similar in the the information they provide. The only functional differences between the VASI and PAPI is that the VASI has the red over the white, the PAPI the white actually goes to the right of the red, and the PAPI offers higher precision (or more glideslopes depending how you look at it a larger airliner with a high cockpit may elect to fly a slightly higher glidepath). The concept is the same though.
A VASI looks like this:

Usage:
As the saying goes,

Red over White, you're alright. (on glidepath)
Red over Red, you're dead. (too low)
White over White, you're out-of-sight (too high)

A PAPI like this:

A regular VASI only offers one glide slope and is designed for an aircraft where the cockpit isn't so high up. However, there is a such thing as two-light PAPI, and a three-bar VASI. So either of them can really be tailored to fit the costs and types of aircraft flying to that airport.
In the case of a four-bar PAPI it is higher precision. Since the PAPI systems uses a narrower beam of light you must fly the glide path more precisely than the VASI to stay on the beam. The PAPI, with its extra lights, forewarns you when you are drifting from the desired glide path.

(source: airportlightingcompany.com)
So one red light would indicate slightly above glide slope, two and two would indicate the normal glide slope, and three red lights would indicate slightly below the glide slope on a 4-bar PAPI. A Three bar VASI works in similar fashion except there are only two glide paths with two reds being the lower, two whites the higher.

Anyone else get frustrated at mnemonics like this? Instead of just remembering that white means high and red means low, you have to remember which situation white rhymes with alright and which situation it rhymes with out-of-sight.

@BluegillPrime If you can remember that white is high and red is low, the latter of which seems rather obvious, both systems can be worked out very easily. The rhymes just make it fun.

The PAPI is functionally different from the VASI in terms of precision (hence the name!).
A Precision Approach Path Indicator (PAPI) uses four sets of two lamps, placed next to each other to provide the pilot with 5 distinct levels of precision (r = red; w = white):

r r r r - below glideslope
r r r w - slightly below glideslope
r r w w - on glideslope
r w w w - slightly above glideslope
w w w w - above glideslope

There's also some degree of fading between red and white.
A typical Visual Approach Slope Indicator (VASI) uses four sets of two lamps in a box formation (two on top, two on bottom) to provide something more like 3 distinct sets of information:

r r
r r - below glideslope
r r
w w - on glideslope
w w
w w - above glideslope

The more vertical placement of the VASI lights is the origin of the mnemonic "white over white, you're high as a kite; red over white, you're all right; Red over red, you're dead".
Operation
As far as operation and installation goes, I'm under the impression that both use the same two-lamp modules, configured with lenses, reflectors, and filters. Presumably the cost to install a VASI is lower, which is why they're much more common; alignment and maintenance is also likely to be cheaper as well.
Furthermore, if one set of lamps in a PAPI is out of service, the entire unit is unusable. If a set of lamps in a VASI is out, the redundancy of the system means that it still can provide some information (albeit information that should be used with extreme caution).


text:
How do I find all the frequencies for my local TRACON?
I'd like to program my hand held radio with all frequencies that I can expect to need around my area.  Specifically, I'd like to find an official list of all the frequencies used by the local Tracon facility.
Looking in the Chart Supplement for the primary airport, I get a list of frequencies for approach and departure at that airport.
If I look at charts for airports around the periphery, I can find other frequencies not listed in the supplement for the primary airport.
My example is Salt Lake City.  Outlying airport is Provo Municipal from which you can reach Salt Lake Approach on 118.85.  I'd guess terrain prevents collocation of the 118.85 antenna with the others.
I'd also be interested to know if there are maps available for ATC sectors within facilities.

http://www.radioreference.com/apps/db/?aid=2958

I visited that site while wanting to program my radio.  Very valuable, however, it misses the example frequency of 118.85

What is your question?

Emphasis added to the heart of the question.  I could also add "Do you know if this list exists and where to find it?"

There are sector maps available, and each facility updates them periodically.  They are published.  My understanding is that sectors are joined and separated frequently, even based on the minute to minute workload.
(I have a set of charts showing High, Low Center and tracon sectors. I haven't found a public source for those charts yet.  Here is a set of maps, but I would like to provide direct FAA maps.  https://ivaous.org/academy/index.php/controllers/procedures-and-organization  )
The chart supplement and AFD are good places for common frequencies, however most facilities have additional frequencies.  They might be used for certain events, emergencies, when VIPs are being handled, during extreme congestion and when there are problems with the primary frequencies.  I do not know of a good source for those backup frequencies, which is published.
There are radio reference listings of ATC frequencies, however, they are not always accurate.  While the may show the station frequency assignments, they may not show the true facility usage.
As an example, here is a compilation of Boston Air Route Traffic Control Sectors. https://en.wikipedia.org/wiki/Boston_Air_Route_Traffic_Control_Center#Low_Altitude_Sectors_2
Addendum:  FAA Facility Transmitting Authorization (FTA) would be the document, prepared for each facility, which would list all the frequencies that facility is authorized to use.  The genesis and use of the FTA is covered in this FAA Order: https://www.faa.gov/documentlibrary/media/order/6050_32b_with_chg_1_and_2_incorporated.pdf

For FAA-owned systems, this number will be the Facility Transmitting
  Author ization (FTA) number assigned by FAA Spectrum Engineering. For
  non-federally-owned systems, this number will be the Federal
  Communications Commission (FCC) license number or “call sign” assigned
  by the FCC.  For military-owned systems, this number will be the
  National Telecommunications and Information Administration (NTIA)
  serial number assigned by the FTIA.

So if you are trying to obtain frequencies for a military base, which may be working with local FAA facilities, you will also need to get the NTIA/FTIA data.
The FTA is public information, but I am unaware of an Order or publically accessible database for the FTA.  When I see an avionics inspector later today, I will ask her if she knows where FTAs can be looked up.

The "ATC Tower and Satellite Airport Communications (TWR)" file is available at the FAA's website under the 28-Day NASR Subscription:
https://nfdc.faa.gov/webContent/28DaySub/2022-03-24/TWR.zip
If you search this file for S56 (the identifier for Salt Lake City TRACON), you will find the following entries:
TWR7S56 118.85  APCH/P DEP/P  25269.*A  PVU ANMUTAH  UTPROVO  PROVO MUNI  40-13-09.000N 144789.000N111-43-24.100W402204.100WCDC CEDAR CITY  ANMUTAH  UTSALT LAKE CITY
TWR7S56 118.85  APCH/P DEP/P  25300.1*A  SPK ANMUTAH  UTSPANISH FORK  SPANISH FORK MUNI/WOODHOUSE FLD  40-08-42.100N 144522.100N111-40-03.700W402003.700WCDC CEDAR CITY  ANMUTAH  UTSALT LAKE CITY

(The entries list very detailed coordinates which unfortunately refer to the locations of the mentioned airports rather than the location of the actual radio antennas.)
The inclusion of 118.85 indicates that the rest of the entries for S56 are indeed exhaustive:
TWR1S56 03/24/2022           ANMUTAH                          UTSALT LAKE CITY
TWR2S56
TWR3S56 120.2                                       APCH/P DEP/P
TWR5S56 RADAR             RADAR
TWR7S56 118.85                                      APCH/P DEP/P
TWR7S56 118.85                                      APCH/P DEP/P
TWR7S56 120.7                                       CD/P
TWR7S56 120.9                                       APCH/P DEP/P
TWR7S56 120.9                                       VFR CD/P
TWR7S56 120.9 ;SOUTH OF 41 DEG LAT                  CLASS B
TWR7S56 120.9 ;SOUTH OF 41 DEG LAT BLW 8000 FT      APCH/P DEP/P
TWR7S56 121.1                                       APCH/P DEP/P
TWR7S56 121.1                                       APCH/P DEP/P
TWR7S56 121.1                                       APCH/P DEP/P
TWR7S56 121.1                                       BASIC RADAR
TWR7S56 121.1 ;NORTH OF 41 DEG LAT                  CLASS B
TWR7S56 121.1 ;NORTH OF 41 DEG LAT BLW 8000 FT      APCH/P DEP/P
TWR7S56 124.1                                       CD/P
TWR7S56 124.4                                       CD/P
TWR7S56 124.4                                       IFR CD/P
TWR7S56 124.9                                       BEARR STAR
TWR7S56 124.9                                       BRIGHAM CITY STAR
TWR7S56 124.9                                       NORDK RNAV STAR
TWR7S56 124.9                                       SKEES RNAV STAR
TWR7S56 124.9 ;300-340 ABV 8000 FT                  APCH/P
TWR7S56 125.7                                       APCH/P DEP/P IC
TWR7S56 125.7                                       FINAL
TWR7S56 126.0                                       CD/P
TWR7S56 126.25                                      APCH/P DEP/P
TWR7S56 126.25                                      APCH/P DEP/P
TWR7S56 126.25                                      CGULL DP
TWR7S56 126.25                                      DZERT DP
TWR7S56 126.25 ;250-300 ABV 8000 FT                 DEP/P
TWR7S56 127.0                                       CD/P
TWR7S56 128.1                                       ARCHZ DP
TWR7S56 128.1                                       DELTA RNAV STAR
TWR7S56 128.1                                       JAMMN STAR
TWR7S56 128.1                                       SEVYR DP
TWR7S56 128.1                                       WAATS RNAV STAR
TWR7S56 128.1 ;160-250 ABV 8000 FT                  APCH/P DEP/P
TWR7S56 128.6                                       BONNEVILLE STAR
TWR7S56 128.6                                       FAIRFIELD DP
TWR7S56 128.6                                       LEEHY RNAV STAR
TWR7S56 128.6                                       QWENN RNAV STAR
TWR7S56 128.6                                       SALT LAKE DP
TWR7S56 128.6                                       SPANE STAR
TWR7S56 128.6                                       ZIONZ RNAV DP
TWR7S56 128.6 ;110-160 TCH ABV 8,000 FT             APCH/P DEP/P
TWR7S56 135.5                                       APCH/P DEP/P IC
TWR7S56 135.5                                       RUGGD DP
TWR7S56 135.5 ;340-110 ABV 8000 FT.                 APCH/P DEP/P
TWR7S56 233.7                                       APCH/P DEP/P
TWR7S56 233.7                                       APCH/P DEP/P
TWR7S56 233.7                                       APCH/P DEP/P
TWR7S56 257.2                                       APCH/P
TWR7S56 257.2 ;SOUTH OF 41 DEG LAT                  CLASS B
TWR7S56 269.625                                     APCH/P
TWR7S56 284.6                                       APCH/P DEP/P IC
TWR7S56 284.6                                       FINAL
TWR7S56 290.3                                       BEARR STAR
TWR7S56 290.3                                       BRIGHAM CITY  STAR
TWR7S56 290.3                                       NORDK RNAV STAR
TWR7S56 290.3                                       SKEES RNAV STAR
TWR7S56 290.3 ;300-340 TCH ABV 8000 FT              APCH/P
TWR7S56 307.05                                      APCH/P DEP/P
TWR7S56 307.05                                      ARCHZ DP
TWR7S56 307.05                                      DELTA RNAV STAR
TWR7S56 307.05                                      JAMMN STAR
TWR7S56 307.05                                      SEVYR STAR
TWR7S56 307.05                                      WAATS RNAV STAR
TWR7S56 316.15                                      APCH/P DEP/P
TWR7S56 316.15                                      DEP/P
TWR7S56 316.15                                      RUGGD DP
TWR7S56 319.25                                      APCH/P DEP/P
TWR7S56 319.25 ;NORTH OF 41 DEG LAT                 CLASS B
TWR7S56 319.25 ;NORTH OF 41 DEG LAT                 CLASS B
TWR7S56 319.25 ;NORTH OF 41 DEG LAT                 CLASS B
TWR7S56 319.25 ;NORTH OF 41 DEG LAT                 CLASS B
TWR7S56 322.3                                       APCH/P DEP/P
TWR7S56 322.3                                       FAIRFIELD DP
TWR7S56 322.3                                       LEEHY RNAV STAR
TWR7S56 322.3                                       QWENN RNAV STAR
TWR7S56 322.3                                       SALT LAKE DP
TWR7S56 322.3                                       SPANE STAR
TWR7S56 322.3                                       ZIONZ DP
TWR7S56 322.3 ;110-160 TCH ABV 8000 FT              APCH/P DEP/P
TWR7S56 335.8                                       CD/P
TWR7S56 353.6                                       APCH/P DEP/P
TWR7S56 353.825                                     CGULL DP
TWR7S56 353.825                                     DZERT DP
TWR7S56 353.825 ;250-300 ABV 8000 FT                APCH/P DEP/P

However you should take that "exhaustive" comment with a grain of salt, because I just looked up my own facility and there is one VHF/UHF pair that is not listed. Strange.

These might help.
https://www.natca.org/index.php/insider-articles/630-facility-spotlight-salt-lake-tracon
https://www.radioreference.com/apps/db/?aid=2237
I use radioreference.com to find ATIS and Tower frequencies at KSFO my home airport.
Hope this helps,
Charlie :)

Please include here the relevant parts of those pages. The pages might disappear and your answer would not be useful anymore.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Variables of Dryden Wind Model
How to determine the turbulence intensities (σu, σv, σw), scale lengths (Lu, Lv, Lw) & Angular frequency (ω0) in the given Dryden wind model to generate wind speeds (u_w, v_w, w_w) of civil jet transport aircraft that experiencing various levels of turbulence?

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What are the relations between Angle of attack and Calibrated Airspeed?
I would like to estimate Angle of Attack using trigonometric relations i.e AoA =  arctan (w/u) airspeed components (i.e u, v, w). When I am considering body 'x' velocity ('u' component), is it necessary to use the Indicated Airspeed? Instead I would like to use Calibrated Airspeed. Is it acceptable? Could anyone please clear my doubt?

Your title is misleading. The question is fairly clear and valid, but what most people will assume from the title is something like "In normal flight, how will change of CAS affect AoA?"

You must use the same quantities for both speeds. Usually, this will be TAS.
As you see, you are going to use a ratio of speeds: w/u. This ratio must be dimensionless, and furthermore, the quantities must be "compatible". For such purposes, you can't meaningfully divide IAS by TAS (or CAS), or, say, altitude by flight level, even though they are both expressed in kt (or ft). They are just different physical quantities.
So, if you "would like" to use CAS for u, you must use CAS for w. How do you get it?
If you are trying to ascertain AoA that way on a real airplane in flight, you'll have difficulties measuring w (body normal airspeed). There is no sensor for it.1 And if you were to imagine some sort of a vertical Pitot tube, its IAS would have very different instrumental errors to the "normal" forward IAS, making these two IASes "incompatible", and you'd need to calibrate it extensively (and thus use CAS).
But the next catch is to check what CAS actually tries to express. Normally, the pilot (as well as the aerodynamicist) is interested in the "flight path" (total) airspeed rather than in the "body forward" airspeed. IAS measures the latter, but the standard CAS conversion may include correction for the AoA, amongst other things. If this is the case, you'd need to use arcsin instead of arctan (assuming you can get "CAS w").
In practice, one uses this formula to obtain AoA when solving equations of motion numerically. In this case, naturally, you have TAS everywhere as the primary quantity. You don't even need CAS nor IAS except for indication to the pilot, and perhaps as an input to autopilot or other equipment if you are simulating it.

1 Note the VSI measures vertical speed (in ground, not body frame), and it's not airspeed at all.

Fine. I have few more doubts. 1) Are true airspeed expressed in x, y, z axis displayed/available on an airplane's computer? If I want to use both the u & w component from TAS, is it readily available during flight? 2) Suppose in the absence/failure of Pitot tube(s), what may be the other way to estimate AoA/AoS?

@S.Prabhu, whether it is an air computer or old-fashioned gauges, you need to start from the sensors: what is actually measured? For most GA airplanes, you only have a Pitot tube and a static port (there are often more than one of the latter, but they are usually plumbed together as one input). This implies that you can only have something closest to "u component of IAS", which you (or the computer) can correct to "u component of CAS", and then with some assumptions about AoA, to "total CAS", and then considering the static data to "total TAS". You can't reliably have AoA from here.

As for your (2), it deserves a separate question. Generally, estimating data from incomplete observations is a big topic in itself. You can't have AoA in this case, but you can estimate it if you make some assumptions, the first of which is your aircraft performance. Say, if you know current weight and power setting, and flying straight & level, you can fairly reliably deduce AoA and CAS for this specific aircraft.

I have seen in some articles that they estimated AOA using gps 'z' & 'x' velocities (represented in body frame). Will it work in real-time applications? 2) I noticed some aircrafts with multiple static holes on the fuselage. In general, how the static pressure input fed to adc? By taking average of all holes or individual/selected inputs considered?

@S.Prabhu, these are all worthy questions that deserve better than be discussed in 500-char comments. For (1), it will be valid only in still air (and that's before we start on the question of accuracy). For (2), differently (I already mentioned the most common arrangement), but if you think you could use the difference between left and right to estimate sideslip, you are right.

One more doubt sir.  AoA estimation using difference of pitch angle and flight path angle (i.e θ-γ). Will it work? Just for a backup when all the redundant AoA sensors are not available/failed...

@S.Prabhu, even this is not a trivial question. "Work" for what? Mathematically, it will "work" in still air. But you can't just disregard wind in most cases. Then, how do you measure the flight path angle? GPS? This is very "slow" data, whereas AoA is one of the "fastest" signals (changes very quickly e.g. in a gust).

I thought to use the formula, γ = sin^-1(RoC/Velocity of airplane) to compute FPA.  I am trying to make an algorithm for failure accommodation of AoA sensors with alternate instruments/sensors available onboard (e.g. Airspeed sensor, gyroscopes) . The ultimate aim is to save the a/c from stall and to land in the nearby airport safely. Could you please share me what are important the points to consider in this case (failure accommodation)? I think your valuable comments may help to improve my knowledge.

@S.Prabhu, please ask it as a separate question. You'll need to list scenario(s)/failures you want to protect from, and your ideas how to work around them.

Ok sir. Can I have your email address?

@S.Prabhu, no, ask here, on SE. There are plenty of competent people here.

Ok sir. Thank you for your patience and replying all my doubts.

Calibrated airspeed is essentially Indicated airspeed with corrections applied, so that should be fine for what it sounds like you're doing.


text:
Why DA40 got some aircraft variants with gust lock on the left seat and some on the right seat?
I noticed that the gust lock position differs for some DA40 aircraft..what are the reasons for such difference in mechanical design?

Is the mechanical design actually different? I.e. are there actually two different models, one for left and one for right? From the answer below it would appear the lock is ambidextrous and capable of being installed on either control stick.

The AFM indicates that the gust lock must be wrapped once around the stick and the rod engaged with the pedals fully slid back:

(source)
it does not specify which side it is to be mounted on although the AFM depicts it on the pilots side there is no requirement for it to be mounted there and it fits fine in the other orientation.

(source)


text:
how to calculate wind speed and direction using flight data, at some latitude and longitude?
I have a data file which has the aircraft state details with latitude and longitude recorded for certain radius.
Now, I'm trying to calculate wind direction and wind speed.
Example:
At lat=41 and long -75 I have the aircraft state details as (flight:AZ9867, nav_heading, mag_heading, track, Ground speed, and TAS)
How can we calculate wind speed and direction?
Do we have to use vector calculation?

"Payload" is not the word you want here.  But yes it's a vector calculation.  It's not clear whether you are asking about what each term represents, or how to do the vector calculation, or both.  Can you clarify that?  (But some of those questions may have been asked already here.)  (Btw, the meaning of "nav heading" is unclear to me.)

I'd say TAS-ground speed

@sophit  That would give you the wind component parallel to the track. By comparing the heading to the track (and considering TAS), you could probably compute the the wind component perpendicular to the track, and then you'd sum the vectors.  Fair bit of math involved, and depending on the accuracy/precision of the data, the accuracy of the result might be limited.  This looks almost more like a physics/programming question than aviation, but the community can weigh in on that.

@RalphJ: well unless you have a test probe installed on the nose of the aircraft, I don't think it's possible to measure any wind direction other than the one from the pitot tube (?)

Possible existing answers include-- https://aviation.stackexchange.com/a/35431/34686, https://aviation.stackexchange.com/a/23435/34686 (note that point #7 refers to a vector subtraction), https://aviation.stackexchange.com/a/81766/34686, https://aviation.stackexchange.com/a/33688/34686

@quietflyer: yes, I do agree that it's a simple vector math problem and I think your links actually answer this question

Btw "course" in third link in my comment above is probably the same as "track" in OP's database.  "Course" can also have other meanings in navigation, but not here.

We can solve this using trig.  Use the Law of Cosines (III below) to identify the wind's speed.  Use the Law of Sines (IV) to solve for the wind's angle relative to the heading.  (Beware that law of Sines will often produce two values and Excel, for instance, defaults to the acute one, which would not be right in this case.)
This assumes that you already have the distance and course between start lat/lon and end lat/lon.  This could get really tricky if you are trying to solve this over a long distance where the spherical nature of the earth means that your ground track and heading a changing continuously throughout the flight.  On a flight from LAX to JFK you will depart on a track of 66° and will be on a track of 93° when you arrive.
That makes wind calculation over a long flight meaningless...  Wind correction is really a zonal phenomenon meaning it's only useful to solve in a specific small segment of a flight.  And thus we skip the spherical coordinate complexity in this illustration.
I'll use my smartboard and avoid an hour spent in MathJax.

I think he's got all the data he needs to instantaneously solve the problem for any given particular lat lon.  It's not like he's trying to deal with measurements of how far the aircraft drifted off course during 2 hours of flying or something like that.  (I think).


text:
Very low, difficult to see, SILENT, Aircraft
I live in Oklahoma, and at about 7-7:30 pm on 11/20/22, something massive flew over my house(going east), bigger than any cargo plane I have seen. It was completely silent, with 3 red lights along the bottom (one in front, middle and rear). Extremely difficult to see, almost blending in to the sky. And moving very slowly. It did not look like it had wings, so I don't think it was a plane. I am prior Air Force, and grew up near Beale AFB, and have seen air craft my whole life. Does anyone here, know what the heck this is?
I thought maybe a blimp, but I have seen blimps before, and those make noise. This made NONE.

Are the lights the only thing you saw of it, or did it have any other visible form?

Silent, slow, and difficult to see = very high up?  Maybe sunlight at dusk reflecting off of the bottoms of a formation airliners approaching an airport?  Anything from area 51 probably would not have its lights on.

there are apps which allow you to fly large numbers of drones (with lights on them) in precise formation by remote control. It could have been three drones, high enough to make their motors hard to hear above background noise.


text:
Why do helicopter performance charts for hover out of ground effect (HOGE) have a "knee"?
Have a look at the following performance charts from Section 5 of Robinson R-22 Pilots Operating Handbook for hover out of ground effect (HOGE). If the temperature and gross weight are known, the hover ceiling of the helicopter can be determined through these charts:

The feature all three charts have in common is a 'knee' - a sudden change in the gradient of the slope. But why is that? Why do the slopes get steeper beyond a particular gross weight?

@AdityaSharma- I suspect a "knee" is a line that increases sharply and then at some point changes slope to something less steep.  Like a leg bent at the knee.

This is conjecture but isn’t there a governor point at which you are maxed out on engine HP and rotor RPM, and any further weight increase must be accommodated solely by altitude loss?  Also, it made me wonder if you’re torque limited in those zones to protect tail rotor effectiveness.

@MaxR: you might be right. Gearbox is also normally torque limited.

The knee is a result of having the engine "flat rated", that is rated to a power limit that is below its potential wide open throttle horsepower at sea level, a very common practice with helicopters. The Lycoming O-360 engine it has is normally rated at 180 HP when used on regular airplanes (wide open throttle at sea level).
To give the helicopter flexibility in altitude performance, they arbitrarily limit its power at lower altitudes to hold back some "in reserve" you might say. That is flat rating.
The R-22 is derated to 131HP at sea level and up to its critical altitude even though its displacement makes it technically capable of making 180 HP. Below critical altitude, you will be less than WOT, observing a redline on the manifold pressure gauge (24.5 inches of MP, whereas WOT will be around 29-30) that sets the 131HP limit. At the critical altitude, you will be at redline with WOT and the engine making 131HP, and HP available declines with altitude above that.
The slope of the curve is the gross weight limitation with altitude for HOGE at a range of temperatures. The steep part of the slope represents the reduction in GW imposed by aerodynamic factors unrelated to engine horsepower. Then you get to the critical altitude, where you are now at WOT to just make the manifold pressure redline, and the decline in power with altitude increases the decline in allowable gross weight.
So, the knee in the curve is the point at which the critical altitude is reached at different temperatures for the particular configuration depicted in the chart.
You also have a kind of unofficial "emergency reserve" available if you really got yourself in trouble below critical altitude, but you have to remember the transmission/rotor are designed for 131 HP and you'll overstress the transmission system and rotor by using WOT in some kind of rescue-yourself-or-die situation that takes the engine past 24.5" of manifold pressure. Not sure how the manual handles that, but owner will be unhappy with you if it affects the life of the transmission/rotor and they find out, but better than crashing.

Very good answer. Maybe you can add that the engine could be derated also due to gearbox limitations and it makes lose some payload at low altitude.

Thanks. Well they'll design the gearbox for the horsepower the machine is designed for regardless. The fact that the engine is derated isn't really relevant. The machine and its gearbox is designed to handle 131 HP.  They could just put a Lyc O-290 in it, un-derated and have the same sea level performance with that gearbox.

Added some extra in response to your comment.

Yes, that's what I meant: the gearbox is "derated" at the same level of the engine.

Yeah usually some margin is built into the entire drive system say if they want in future to upgrade to the 200 hp version of the O-360 and flat rate THAT to say, 150 hp, without having to beef up the drive system/rotor. Depends on whether they can tolerate the weight penalty of "extra meat for future needs".

Another bonus is, an O-360 that never has to make more than 131 HP is so under stressed it could probably run 4000 hours to overhaul. When it gets to the 2000 hr rebuild after a few years of steady regular operation, the inside of the engine will be pristine.

So the knee is the point where the performance limit changes from RPM limited to power limit.

Not RPM limited. These things run at or near redline RPM full time. Just the point at which it takes wide open throttle just to get 24.5" of manifold pressure, the limit for the derated engine.  So power available is constant up to the critical alt, b/c you can keep opening the throttle to maintain 24.5". Above critical altitude, wide open throttle is LESS than 24.5", declining as you go up.

There is something in section 7-4 about using the collective above a certain altitude to maintain rpm once the engine is maxed out on power.
The steep part of the weight vs maximum hover altitude curve  indicates performance is based on the (collective) rotor pitch at an Angle of Attack matching the engines power output while maintaining constant rotor rpm.  The decrease in rate of lifting capacity is consistent with exponentially increasing Drag and increased rotor interference as rotor AoA increases.

at the "knee" the engine is at full throttle and the rotor blade is at optimal AoA (maximum Lift/Drag ratio) at maximum allowable RPM

Above a certain (density) altitude decreasing Wide Open Throttle engine power output at constant rpm means rotor blade pitch is further reduced to maintain rpm.  But rate of loss of maximum hover altitude as weight increases is not as steep because rotor Lift to Drag ratios are more favorable and rotor interference turbulence is reduced.
Another way of looking at the "knee" concept would to mount a large rotor on a tiny motor.  It could only spin a constant rpm up to a limited AoA before the engine simply could not give any more power. The point at which AoA cannot be increased any more is the "knee".  Really a wall.
Now, take a tiny rotor on a very large motor.  AoA could be increased until until the blade stalled, resulting in a straight line of lift increase, then a sharp decrease.
Practically, matching helicopter motors with rotors tries to avoid both extremes, erring on the side of limiting engine power.  Under conditions of manifold pressure limitation or "hot/high" air density, the knee will show up on the graphs.


text:
Why aren't passengers trained and evaluated for operating emergency exits in commercial aircraft?
The passengers sitting on the overwing exit doors are expected to operate the doors in times of emergency, with only a verbal instruction. Is that enough?
Q. Why can't we have a designated area in the airport where a setup of emergency doors are kept and the passengers are trained and evaluated there before being alloted the seats on the emergency row. It can be short 2 min course, where passenger actually opens the door and puts it aside.

There have been instances which highlight that proper training is required for passengers Soource:

Passengers will frequently open an exit as soon as evacuation begins which may result in them evacuating into danger. This occurred in the Ryanair engine fire in Stansted Airport in 2002, where passengers evacuated themselves onto a burning wing, despite airport fire services personnel shouting at them to return inside the aircraft and evacuate via a usable exit. Source

Another hazard in the use of overwing exits is them being improperly opened (usually a result of passengers in these seats not paying attention to the verbal briefing provided pre-departure, or not observing the opening instructions on the safety card and on the exit). The majority of overwing exits involve the passenger physically removing the hatch from its frame and disposing of it outside on the wing without blocking the exit. Research conducted at the Cranfield Institute in the UK Source

Above situations taken from Wikipedia
https://en.wikipedia.org/wiki/Overwing_exits#Hazards

In my country some airlines sells those seat for extra money, because they have extra leg room. And usually elderly people buy those seats. Sometimes there are people barely able to walk sitting next to emergency door. I agree what youre saying, we should check if those passengers able to open that doors in an emergency.

I honestly don't think the "2 minute training" (or 2 hour) would matter, people do a lot of weird stuff when they panic. The physical act of opening the door is not difficult, the training to think calmly in an emergency is something that would take weeks or months.

Even trained crews panic, like this crew screaming and yelling after a loss of cabin pressure.

If the exit row passengers didn't go to the training (not enough time, spent it eating/drinking/shopping, didn't know they were supposed to, just didn't bother, etc...), what would you do? Leave the row empty (bumping passengers if the flight was full)? Is that really an improvement over untrained passengers?

@AliErdem Are you sure the offer emergency row seats who are not able to walk properly? That is a big security risk IMO.

@RonBeyer Even though I agree to your point, but doing something for the first time in a panic is wayy worse, than having done it earlier. That is what trainings are for.

@ZachLipton I've seen passengers sitting on emergency rows who were un-cooperative during their briefing. The cabin crew were quick to change their seats.

@Firee Use of the doors is pretty rare. Doesn't sound worth the effort.

Former TWA here.
Emergency exit seats used to only be given to passengers who are ready, willing and able to assist in case of an evacuation. They could not be assigned in advance, the passenger had to be asked and evaluated in person by the boarding agents, who would have been held responsible if a not-able passenger was to be seated in any of those rows/seats. This used to be the rule, it was deemed that safety was more important than selling seats.
Maybe the rules have changed since?

I think I've seen, when buying a ticket online, popups that have you click a checkbox confirming that you can operate the emergency exit.  I guess that fulfills the same (legal?) requirement, but probably much less effectively.

In my experience, the "willing and able to assist" requirement is still checked (1) as a popup when selecting an exit seat online, as Maxpm noted; (2) by query from the agent at the boarding gate when the boarding pass is scanned; and (3) by query from a flight attendant on board after briefing the exit row passengers. All of these, however, are mainly a self-evaluation by the passenger, who just has to say (or click) "yes". Some further basic checks can be done either using ticket data or visually, such as that the passenger is at least 15 years old (in the US). ...

... But none of that, IMO, really addresses OP's question. OP was asking why there isn't a practical evaluation/training using a physical replica of an exit row. OP already noted that exit row passengers are given "verbal instruction", which is pretty much what this answer refers to as "asked and evaluated in person by the boarding agents". So I find this answer off-point unless it is explained further why the at-a-glance "evaluation" that does occur is equivalent in power to the evaluation that OP actually proposed.

Training is indeed verbal, plus there is an emergency instructions pamphlet at emergency seats. Please note, not all emergency exits are operated by passengers. At TWA boarding agents were instructed to choose obviously able passengers, such as non-revenue who has been trained by the company, military, EMT, mainly people who in case of an emergency, would jump in action instead of panic. To see how it worked, please refer to TWA-843 burned down after aborted takeoff at JFK (L-1011) with full load, all were evacuated under 90 seconds: only 2 minor injuries. That proves it works.

In short because accidents do not happen. Airline accidents are incredibly rare, so training the average passenger for them simply does not make sense.
It's easy to overestimate the number of accidents by looking at news articles about a crash. But look at the following picture:

You can hardly see Europe. There's a bit less in USA, as it's night there at the moment this screenshot was taken. But in short: none of these flights crashed. It's thousands of flights per day, and accidents typically happens once or twice a year. Training for it doesn't make any sense.
Training for how to get out of your car after an accident would make much more sense - yet very few do it...

Hard to see Europe, indeed. But objects on map are smaller than they appear...

How much training, for how long, would you require?
Do you realise how many different door designs there are, there are hundreds? The facility would have to be massive.
And who'd pay for it? Another $100 surcharge on every ticket just to pay for the facility and instructors would probably be called for.
I'm sure there are more reasons why this is not a good idea.

I under stand the complexity involved if one were to be 100% precise. However my though was  simple, just one or two generic emergency doors, maybe A320 and B737 types. Just the physical act of operating the emergency door would give an adrenaline rush to the passenger, along with clapping by fellow passer byes.

@Firee that'd then not be training, but entertainment. Wouldn't help at all in a real emergency. But if you think it can be made commercially viable, I'm sure there'd be airports willing to rent you some space in the departure lounge to set up something like that.


text:
Area under curve lift vs angle? Any significance?
I'm looking at AOA vs Cl graphs and comparing low AR vs high AR curves. What does the area under the curve mean? Thank you

What does the integral of the Cl wrt alpha mean…

I have never seen any significance to that area.
There certainly is significance to the derivative of that curve -- the lift curve slope.
The area would have 'units' of lift coefficient (dimensionless) times angle.  While that is certainly not a proof, looking at the units gives a clue if there will be meaning.  Consider Power vs. time -- power*time is energy -- the area under a power vs. time curve is energy.


text:
Why are taildraggers considered more rugged than tricycle?
Not trying to get into the age old debate of which is better. From mechanical perspective, all landing gears are just long cantilevered beams. With a greater instability, one would expect tail draggers to not be preferred for a rough ground surface.

Only thing I can think of has nothing to do with the gear, but keeps the prop further away from the ground. Nose-wheel airplanes have very small prop clearances while tail draggers have good prop clearance.

Related: Why tail wheel rather than tricycle?

If you land on a grass runway and drop the little wheel in a gopher hole, what happens next?  I'd expect a nose dragger to tumble upside down.

@DanPichelman do you mean a headstand?

@dalearn - headstand, nosestand, really bad day

Non-retractable taildraggers can be fitted with large, low pressure tires which are much better on soft and/or rough terrain. Tire pressure strongly correlates with the possible loading and smoothness of the airfield, and once landing gears became retractable, tire size had to be reduced to fit them inside the smallest gear bay possible.

@Peter Kämpf: Is there any reason why non-retractable tricycle-gear aircraft can't be fitted with the same large tires?

@Dan Pichelman: Generally you keep weight off the nose wheel as long as possible, at which point you should be going too slow for dropping the nosewheel into a hole to cause a nose plant.  Not that it's not likely to ruin your day. but I've landed on a good many grass & dirt strips (and a few dry lake beds & such) with a Cherokee.

@jamesqf: agreed, but which would you prefer to land into a field with the furrows plowed perpendicular to your course?  That would be a bad day no matter what but I suspect it might be worse with the little wheel in the front.

@jamesqf: Drag. Three big tires cause more drag than two.

@Peter Kämpf: Just as 3 small tires cause more drag than two.  And wheel pants reduce drag, but are seldom seen on planes that spend much time on grass or dirt.

@jamesqf: Additional advantage of trikes for soft-field operations: you can (I would assume, at the very least) put a bigger tyre on the nosewheel than you can on a taildragger's tailwheel, thus increasing contact area and further decreasing ground pressure relative to the taildragger.

To expand/clarify upon the other answer, bush planes and the like use tail wheels because imagine if when landing on a rough runway, your little tiny nose wheel hit a rock.  That could be catastrophic.  However, on a taildragger, the big main landing gear can go over some obstacles, and the little wheel on the back won't pose a big problem with small rocks and such.  The little wheel will just bump over the obstacle, instead of possibly skidding on it and throwing the whole plane off.  Effectively, the plane is dragging the tail wheel over the obstacle vs. pushing the tail wheel into the obstacle.  So, that's why I think that tail draggers are preferred for a rough surface.
Feel free to correct me if I'm wrong... I've never really flown a plane in my life.  I'm just an enthusiast.

this is not "another reason", but a consequence of what I wrote. A nose wheel is pushed down into the rock, while a tail wheel is pulled up and over.

True, but I expanded upon a specific part of your answer.  If you feel I plagiarized your answer, I will delete this answer and relegate it to the comments section.

Don't worry. I do not care about reputation points. But you why don't you write that you want to expand on the other answer, or add another angle, etc. That would be easier for the reader.

Counterexample: Twin Otters have trike gear and yet serve as excellent bush planes.

Let's assume that the main wheels are more sturdy than nose or tail wheels.
When the plane encounters an obstacle on the runway, the inertia of the plane will push the nose down and lift the tail up.

the force on a nose wheel will increase
the force on a tail wheel will decrease

Additionally the tail wheel is further away from the COG than the nose wheel, resulting in less force needed to lift the tail compared to the force needed to lift the nose.

Thanks for your thoughts. Your last comment makes me think that the less force needed to lift the tail would make the airplane more likely to flip on its back.

@SP00N Yes, but this is about the pivot points. When only the main wheels are down, the nose wheel plane lifts its tail easier because its main wheels are further backward than the taildragger's main wheels. As soon as the nose wheel is down, the nose wheel plane cannot rotate forward any more without breaking the front gear. When the taildragger rotates forward, the next parts after the main gear to contact ground are propeller and cowling...

Wouldn't hitting an obstacle bounce the nose up, not down?

Obstacle means the wheels cannot roll over. The wheels are stopped, not lifted. The plane's center of gravity has a forward momentum and will start rotate around the wheels' axes. This turns the nose down.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How do different training paths to multi-engine rating compare?
I am going to be initiating my private pilot license shortly and wanted some feedback and comments with regards to utilizing single engine (minimum) and multi-engine to obtain my multi-engine rating quickest.
I know that most people start and stay with a small single engine to obtain their license. Then start putting some time in (later) to get their multi-engine.
However, I am thinking that after a little time in a single—doing most of time in a twin (even though the cost is significantly more) because my time in the twin will apply later on towards the multi-engine will be more efficient.
If it's been objectively compared before, I'd like to know the different paths taken, and which is better when?

What country are you wanting info for?

This question could be better answered if you clarified your career goals, and country, and whether time or money spent is more important to you.

At the end of the day the choice might be made for you, as many flight schools will not let you fly their multi-engine aircraft until you have a private licence.
People almost always begin on single-engine aircraft not just for cost, but also for efficient training. Multi-engine aircraft are complex beasts that require a lot of theoretical knowledge and practical skills to fly safely. If you're trying to get a grip on all that at the same time as just learning the basics of flying, you'll no doubt struggle and have to repeat lessons. Your first solo will definitely come considerably later, and many would agree that you learn the most about flying when you're up there by yourself.
Granted, the more multi-engine time you can get, the better your future job prospects (generally speaking). Just remember that doing your early training like that will increase the time it takes to get your ticket. For a qualified pilot, a multi-engine rating only takes 10-15 flight hours of training anyway, so to me there isn't a huge advantage in getting one before your private licence.

How do different training paths to multi-engine rating compare?

Your question is very broad and the answers will possibly be heavy on opinion, and this website much prefers objective answers. That said, there are objective items that can be considered.
For example, let's say you're away from the airport and your instructor simulates an engine failure. If you're in a single-engine plane, the proper response is totally different than if you are in a twin-engine aircraft.
Or, let's say your instructor feels you need to experience landing on a grass strip in the mountains with an uphill-downhill runway. When I was instructing, I always took my students into such a strip, McKenzie Bridge State Airport in Oregon. However, I never took a multi-engine student there as it's not a strip you would generally take a twin into.
My point is, if you abbreviate your single-engine experience, you may be foregoing useful training and education. Perhaps, then, your best course of action insofar as when to switch to a twin is to do so when you feel you have learned from your single-engine flying all that can be reasonably learned that is unique to single-engine flying as opposed to multi-engine flying.
You'll only know what that point is when you actually reach it, and to some degree what you are planning to do with your license affects that.


text:
How do hot air balloons navigate?
Hot air balloons are quite obviously carried along with the wind, so how can they be navigated?  Choosing the correct launch point relative to the desired landing point would seem to play a major part of this, but logic states that it is not always possible to choose the best possible launch and landing points.
Can a hot air balloon be "steered" by increasing or decreasing altitude, hence selecting just which wind it will be subject to?  Does the direction of the wind vary with altitude, and if so, how much?  How do balloon pilots know how the wind will vary at altitude if that is the case?

Welcome to aviation.SE!

@Pondlife, thanks.  I've been lurking here for years though.

Okay, given that wind direction does appear to change with altitude, how can a pilot know/predict what that change will be without using online resources?  I.e. what is happening in the atmosphere to cause this change in wind direction?

Not much unless the launch test balloons and observe to see what they do. Most of the time online info is better... which is why they use it.

How much it changes if at all depends on where you are and how high you want to go, some places the wind has too little to no change to "navigate" you just keep going until you fly over a landing spot.  sometimes you can dip down into a valley and the cooler slower wind will be draining down and can get some steerage.

there is a reason why ABQ is the balloon capital of the world, between the mountain blocking and morning drainage, you get lots of steerage, but it is in no way guaranteed, you use weather reports, and release a latex balloon before flight to watch what the winds aloft are before flight

They use the fact that the wind veers when going up, that is, changes direction clockwise, and backs, changes direction counterclockwise, when going down. This is largely (there are a lot of other factors but this is the one we are interested in) due to the tendency of air in the atmosphere to spiral away from high pressure toward low pressure (spiraling because of Coriolis effect; what makes draining water swirl - clockwise away from high pressure in the north, counterclockwise in the south).
Away from surface influence, the air spirals nearly perpendicular to the "slope" of the high pressure (think of it as a mound of air that wants to flow toward and fill in adjacent low spots but Coriolis effect makes it swirl as it does so).  Closer to the surface, friction effects inhibit some of the Coriolis effect make air flow a bit more directly "downhill" toward the Low you might say, the more the lower you get.
For a high pressure area in the northern hemisphere where there is clockwise circulation as air flows away from the centre, this means the wind is more directly away from the center of the high at the surface than when higher up, where Coriolis effect is stronger and it "swirls" more.  As a result the wind direction will change clockwise as you go up.  And vise versa going down.  In the southern hemisphere where the circulation patterns are reversed, it'll be the opposite; veering going down and backing going up.
This gives the balloonist some control over ground track by choosing a specific altitude that gives the desired track.  Obviously, the amount of veering and backing being fairly small between the surface and 5000 ft, maybe 10-30 degrees, means the balloonist still has to pick a launch site that is upwind of destination.  A launch site that is directly upwind based on the wind direction at, say, 2000 ft will give some left and right directional control by going higher to change track to the right, and lower to go left. The balloonist also has to account for wind backing during the descent to landing and allow for that in the "cruise" portion of the flight to get the best position for the final approach.  Beyond that, it's control the altitude with the burner and hope for the best.

Please add some words about Coriolis force and friction close to the ground to give a credible explanation why winds change with altitude as they do.

Well that's a whole new question isn't it.

Why? That is the physical reason for the changing wind directions, and as your answer stands, it just claims how they change without presenting proof. I thought that would be a nice addition.

Is the direction change with altitude hemisphere-dependent?  If so, perhaps you should specify whether you're answering for North or South half of the world?  (Your profile has no location, so I can't even guess).

Ok well I gave it a shot, trying to avoid too much esoteric weather jargon.

That is exactly how balloons “steer” - they pick an altitude where the winds aloft are going the way the pilot wants to - assuming winds at an available altitude exist where the wind is what the pilot wants.
Many balloonists use Ryan Carlton’s wind tool found here to get a good idea of local winds: http://ryancarlton.com
NOAA has a website called RUC soundings that presents actual wind and temperature aloft data from balloon launch sites and then interpolates for locations in between: https://rucsoundings.noaa.gov/
While the pilot can plan a flight and do amazingly well to get near to where they wanted to go, at the end of the day, a balloon, as an unpowered aircraft, is at the mercy of the winds.

Balloonists use spit and little balloons to track the variations of wind direction with altitude.
Before launch, the balloonist will release a small helium balloon and note which direction it drifts as it ascends.
After launch, the balloonist will work up a goodly gob of spit and discharge it overboard from time to time, noting which way it drifts as it descends.
Such measures are limited in accuracy and in the presence of a strong wind, the balloonist has not much choice about where to go. The ground crew in the chase truck follows the flight and knowing the flight's intended duration, will find a convenient location along the balloon's track and park there- thereby defining the balloon's "destination".

Shaving foam or whipped cream is used by the less saliva-oriented pilot.

That seems like a practical way to do it!

@Sanchises Whipped cream has the disadvantage that your passengers might eat your navigation system.

@DavidRicherby: The counterpoint is the disadvantage to the ground crew using the spit method.

@dotancohen So shaving foam seems the safest. Most passengers will either be well-enough groomed that they shaved before boarding, or scruffy enough that they won't feel the need to shave during the flight.

@DavidRicherby I was going to suggest that shaving foam has the disadvantage that your passengers might end up better groomed while you are trying to figure out how to land.


text:
Is a broadhead arrow fired from a bow or crossbow a significant threat to the safety of a civilian helicopter?
Let us suppose that we have a civilian helicopter, such as an MD900, Bell 407, Robinson R44, Bell 47 or another common civilian design, hovering or flying slowly just above treetop level (50-70m).
If a person on the ground nearby was to shoot at the aircraft with a broadhead arrow from a bow or a crossbow suitable for hunting big game, would the arrow pose a significant threat to the aircraft or its occupants?  Would it be possible to hit the pilot through the canopy, or cause a failure that would lead to a forced landing or a crash?
If it was known beforehand that the aircraft would be used to pursue a fugitive armed with a bow or crossbow, might the pilot reasonably refuse or fly in a different manner than if they were performing a mission such as search and rescue of an unarmed person?

This is possibly a question where there needs to be thought on 'do we want to answer how to shoot down aircraft on the site'. For what it's worth some napkin math suggests the arrow will have lost just enough kinetic energy to combine 'will get you attempted murder charge' and 'not actually do anything'.

There was at least one incident in the Vietnam conflict where an American helicopter was brought down by a bolt from a large, powerful crossbow fired from the ground.

@ZeissIkon - "A large, bamboo crossbow device used by Viet Cong forces during the Vietnam War, 29th November 1965. Armed with an eight-foot long spear, the device was positioned near a jungle clearing, and was intended for use against low-flying or landing helicopters. (Photo by Keystone/Hulton Archive/Getty Images)" image here

@MichaelHarvey Yeah, I'd call that more of a ballista than a crossbow

The Bell 47 might be susceptible due to its exposed engine. I'm hard pressed to believe any bow or crossbow could pierce the canopy under the stated conditions. Maybe, maybe, the arrow can minimally pierce the shell if it hits it at exactly the right angle. Which, due to the wash, would be Robin Hood miraculous. Now, a helicopter with a turbine input... it'd be one whale of a lucky shot, but that might do it.

I'm assuming you're not counting the "startle the pilot and force them to make a mistake" issue as a significant threat?

"Plunging fire" may be interesting. Fire close to vertical so bolts drop through rotor disk - and maybe into canopy from above. || I suspect tail rotor strikes may prove more exciting than others suggest.

@bta If that's a significant threat, then yes, that's included.

I suggest you watch the documentary "Avatar".

@GrimmTheOpiner. Or the older ones, Rambo 2 and 3

Yes it could pose some risk to the pilot and machine. As this video shows, arrows and bolts can easily penetrate acrylic plastic and thin metal skins if the range is close enough.

So an arrow or bolt could penetrate the side window/door skin of a light civilian helicopter and cause, at minimum, a threat of injury to the pilot if fired from, say, less than 100 ft away. They often fly with the side doors removed, and in that case the pilot is as vulnerable as someone sitting in the open on the ground if the shooter is accurate (and lucky, as the rotor wash would affect the flight path somewhat and throw off the shooter's aim).
Less likely but still possible, an arrow/bolt fired from close range could penetrate a fuel cell/tank that was integrated with the outer skin and cause a leak, although this wouldn't bring a machine down right away.
Or an arrow/bolt could penetrate the cowl around the engine, and by chance cut a fuel line and bring the machine down, so that's a possibility if not a very high one.
As far as the rotating machinery goes, the item that would be vulnerable to an arrow or bolt would be the tail rotor blades, so a hit on a tail rotor blade could likely have an effect similar to throwing a rock through it, and cause enough deformation damage to a blade to set up a vibration.  A pilot sensing a significant vibration through the anti-torque pedals would immediately fly away and land in a safe spot nearby to inspect the tail rotor.
As as far as the main rotor head and blades go, the effect would be mostly scratched paint.
If it was me flying the machine, and I knew there was somebody with a high powered crossbow or compound bow right below me under a tree canopy, I would certainly be a bit nervous about an arrow or bolt coming through the side door and sticking me, or though the chin bubble and getting me in the legs.
If I was the shooter, I'd be aiming for the pilot for sure, and probably adjust my aim upward to account for the rotor downwash. A hit would be mostly luck though although it would certainly motivate the pilot to leave the scene at minimum if the tip of an arrow was stuck in his/her arm.

And further to John’s excellent points, distance is your friend.  (a) That arrow is bleeding off a ton of speed and is losing penetration energy at the square of that reducing speed.  (b) the linear width of the pilot is decreasing in a linear relationship with altitude, and his/her area is decreasing with the square of altitude.  (c) In law enforcement, most departments would not allow the pilot unlimited discretion to pursue a suspect at low enough altitudes for it to be a problem.  It is neither safe nor tactically advantageous.  You are in the coffin corner for most helicopters.

For those watching the video, the compound bow firing a broadhead is six minutes in, while the crossbow firing a broadhead is at 11 minutes.

Would it be possible to hit the pilot through the canopy, or cause a failure that would lead to a forced landing or a crash? The OP states the helicopter is a minimum of 50m (164 feet) in the air. That's probably 4-6 times the distance shown in the video. No rotorwash, hitting a stationary target at right angles vs. a moving target vs. the curves of the helicopter's shell. This is an interesting video, but I don't believe it establishes the credibility of your answer.

The effective range of a high powered comp bow or x bow is about 40 yds and if a machine is passing banked over enough to provide a reasonably perpendicular surface, it will still easily penetrate the 3/32" or 1/8" acrylic you typically find on side door windows with the right head.

@MaxR: If we factor out flight time: Energy loss with is linearly proportional to height (in a vacuum; with air resistance there are extra loses).  An arrow going upwards is trading kinetic energy (1/2 m v^2) for gravitational potential energy (m * height) at a 1:1 ratio because energy is conserved.  (Again, not counting air resistance: average_drag_force * flight_distance = energy lost to drag, where drag force is a function of speed, from the definition of work = force dot distance.)

@PeterCordes But gravitational potential energy has “left” the system of analysis, so to speak.  Yes, it will be converted into kinetic energy as the projectile falls to the ground, but it is lost in the sense that it is not available to the penetration of the helicopter.  By definition the arrow that missed and is falling to the ground no longer participates in our helicopter threat system.

@MaxR: Yes, precisely.  That's why I said (kinetic) energy loss is linearly proportional to height.  Instead of reasoning about speed squared, just do height.  The extra discussion was for the benefit of future readers wondering about how energy worked.

@John K, Two points.  I know bow archers that get consistent hits at 70 yds.  But also, an arrow fired vertically, just as a bullet, will have significantly greater effective range than horizontal.  Shorter absolute range, but greater effective range which is a function of circular error probability.  The largest sources of error in long range trajectories are range range error and velocity deviation because in an arced trajectory these both equate directly to vertical error.  As a trajectory approaches vertical, the range/velocity mis-estimation error approaches zero.

@JohnK In the UK, competition shooting often extends to 90m / 100yds for "recurve" bows and longbows. Compound bows are usually limited to 70m - no idea why - but at that distance even a novice would expect to hit the gold, ie a 20cm diameter ring with every shot, and even a longbow would hope to hit the target (122cm) every time.

There are several factors to consider, for instance the type of bow, type and weight of the arrow, etc. A hunting arrow from a strong compound bow has an exit velocity of about 90m/s (this is on the high end but possible for a strong and practiced archer). Gravity's acceleration is 9.8m/s2, and using a projectile height calculator that arrow shot straight up will go up to 413m before stopping. That's over-simplified as it does not take into account air resistance (which is very significant), but it still shows that at tree-top level, call it 50ft or 15m, the arrow will still be going pretty fast. I'm going to call it 35m range though, as it's unlikely someone would be shooting straight up, and at that range an arrow with that much power will go into the trunk of a tree.
As for accuracy, a good archer in decent conditions at 35m range would be able to group all arrows within the size of a small plate.
So from an energy perspective as long as the helicopter is not too high up or far away an arrow could theoretically cause damage, and a good archer could theoretically hit it. Aircraft are built light, much of the skin is just thick enough to provide protection from the elements, not armored in any way. That doesn't mean a shot is going to penetrate though, unless the arrow is relatively perpendicular to the surface the arrow will just glance off. If it does penetrate, or the door of the helicopter is open, the arrow could injure or kill someone inside. As for damaging something mechanical that would bring the helicopter down that's pretty unlikely.
The biggest problem I would see with this is the turbulence and power from the rotor's downwash. If I remember correctly the speed of a rotor downwash is the equivalent to a 30mph wind, and is a roiling mass of air, so it's going to disturb the arrow's path in ways hard to predict. You aren't going to hit what you're aiming at, you'd have to make a lot of adjustments and hope.
So, it's possible, that a good archer with a powerful bow and the right arrows to hit a hovering helicopter at treetop level, but unlikely that it would hurt anyone or cause major damage. If I was going out to look for a hostile with  bow it would still be prudent to stay a bit higher and not hover for too long.

But how long will the arrow spend in that downwash?  Sure it will play havoc with the flight path, but it won't have long to act on the arrow.

The disturbed air starts before you get to the rotor, so probably half a second to a second. It doesn't sound like much but it will be enough to throw off your aim.

The downwash will certainly take away your bullseye but that doesn't ensure a miss.

I'm not saying you can't hit the chopper @LorenPechtel, you probably can. As you say, it will take away a bullseye, but if you intend to hit someone inside that's what you need to hit.

Many years ago I found a sniper's handbook at a bookstore (I don't remember the title). Admittedly it was for rifles, but it had a section on targeting vehicles: The tail rotor hub is vulnerable, even on most military helicopters with armoured cockpits and protected avionics/drivetrains.
Engine intakes will always be vulnerable to FOD (foreign object damage), particularly the civilian models which don't have covers over the engines (the MD902 in your first link only has a mesh screen over the intake). Getting a metal (or even stone) arrowhead into the engine wouldn't be very good for it and could potentially cause a loss of power resulting in a forced landing.

In turbofans, enormous centrifugal forces naturally push air contaminants away from the center.  Thus anything like sand or even rain water end up going through the outer part of the fan bypass and can't get into the compressor core.  Turboshaft engines (turboprops and helicopter turbines have no bypass, so they almost always use a centrifugal or inertial separator ahead of the compressor.  The heavier the object that makes it into the intake, the more effective the separator is.  It often looks like the engine is right behind the intake but actually follows a convoluted path to the engines.


text:
What is the difference between Maintenance planning document(MPD) and Maintenance Manual (AMM)?
Could someone explain to a layman, relatively new to Aviation maintenance, the difference between the Maintenance planning document(s) (MPD) and the Maintenance manual (AMM). Is one a subset of the other (MPD --> AMM). Both are provided by the OEM when a new aircraft is delivered to an operator?
Where are the work packages, job/task cards associated with any aircraft maintenance activities derived from? Is this information coming from the MPD with the AMM only serving as a reference to any specific steps? Could someone with the expertise explain this? Thanks.

Both MPD and AMM are provided by OEM.
MPD includes all the tasks (only task numbers / descriptions) needs to be performed with their intervals, required manhours, etc.
AMM is maintenance manual which is describing basic working principle of the systems, how to perform the tasks listed in the MPD.
To summarize roughly,
MPD serves the purpose of planning of tasks.
AMM gives detailed information on how to perform the tasks step by step.

Thank you for the response. Very helpful to get a high level understanding.

The MPD is the maintenance roadmap you might say, for the basic Maintenance Program agreed between the OEM and the regulator, that defines the inspection and test requirements to maintain airworthiness.  It'll include a listing of which inspections and tests have to be done at Routine, A Check, C Check or some time interval, etc.
Inspections means things like zonal inspections and detailed visual inspections, that sort of thing.
"Tests" is generally Operational Tests (does the system work or not) and Functional Tests (does the system meet a performance standard - Under MSG-3, you can run systems indefinitely as long as they meet some performance or measurement limit). Inspections are generally Detailed Visual, close inspection of specific parts or a system, or Zonal, general inspection of a defined area.
The procedures you use on the floor will be Task Cards, that have their own manual that has all of the Maintenance Program test and inspection procedures, with a reference numbering system tied to the MPD.
The AMM is much broader, and is the technical source for the Task Card content and all other procedures on how to work on the plane, like Removal and Installation procedures, Return to Service tests, etc.  When you peruse the AMM you'll find that some procedures have Task Card and MPD references at the start, and some are stand alone procedures with no external references (mostly Removal and Installation).
Task Cards will typically be an AMM inspection, ops test, functional test, or special test, as required by the approved maintenance program, and more or less extracted word-for-word from the AMM, so that when you are assigned a scheduled maintenance task you just need to be provided the Task Card and work from that.
The procedure you follow will be a verbatim extraction from an AMM procedure which is the foundational technical source for most maintenance on the airplane.
So, if you have to go beyond the content of the Task Card because you have to replace something that failed a test or inspection, or if you are doing something to the airplane not directly related to the maintenance program, like resolving snags, or doing an ops test or functional test for troubleshooting purposes, that's when you have to dig out the AMM procedure itself.

Thanks for the detailed information. Its much clearer now.


text:
What was the flying instrument panel layout type behind the yoke onboard the Antonov 225?
Was the layout of the instrument panel considered to be basic T type and what were the instruments on it?

Does this image below shows the correct instruments on the instrument panel?

I understand you meant An-225. Its cockpit was almost identical to that of An-124, apart from the engine group.
Basically, yes, it's almost a standard T. The only difference is that on the right, apart from the altimeter, there is TCAS (+transponder) - the black square digital display. Vertical speed is further to the right (a bit higher for the captain).

ASI/Mach | AI  | TCAS | VSI
---------|-----|------|----
         | HSI | Alt  |

(Source: An-124 manual, but it's difficult to quote it directly: apart from being in Russian, it is laid out per sub-system, and there is no united dash description).


text:
What are these points in this drag curve ("Lilienthal'sches Polardiagramm")?
If it matters, this is taken from EASA PPL questions by AustroControl.
So far I think the points correspond to the following situations:

1 Inverted Flight (confirmed)
2 ???
3 Maybe Best Gliding Angle???
4 ???
5 Slow Flight (confirmed)
6 Maybe Stall ???

Look at what's unique about each point. (2) generates zero lift. (3) is the point of minimum drag. (4) is the point of maximum lift/drag ratio. (5) is the point of maximum lift. Does that steer you in the right direction?

Thanks. Here's what I think without opening the book.

2 stall
3 max endurance
4 max range speed
-6 this is now even more confusing

6 is the stall. 2 is just the zero-lift point (like NASA's zero-g Vomet Comet would want - completely unloading the wing). 4 is a very important point to understand: best glide AoA (= speed more or less)!

Ok, so 5 is CL Max, or the Critical AoA?
Then does this mean that the Blue line then is the Total Drag and in the lower part shows Induced Drag and then in the upper half Parasitic Drag takes over?

Right on with (5). It's the other way around with drag: parasitic drag dominates at the left side (at (2) it's all parasitic drag since there's no lift to induce any drag). Induced drag dominates to the right at high (positive or negative) AoA.

Keep in mind that each point just gives the instantaneous Cl and Cd, or in the nomenclature of the graph, Ca and Cw, of some particular Angle-of-Attack.   So, point #1 could certainly be obtained in non-inverted flight, if the pilot shoved the stick forward suddenly.  The G-load (basically another word for the lift vector) would be negative.  This is not compatible with a steady-state condition, unless the aircraft is inverted, but that doesn't mean that the aircraft must be inverted to experience this part of the flight envelope.

Re " So, point #1 could certainly be obtained in non-inverted flight, if the pilot shoved the stick forward suddenly." -- most safely demonstrated starting from a steep climb.  Just as you would demonstrate a slightly less extreme version of the maneuver, bringing the wing to the zero-lift angle-of-attack, and bringing the G-load only to zero, rather than to a - value. If you attempt to demonstrate either of these maneuvers from normal level flight, in most aircraft you will end up with the aircraft very steeply nose-down and the airspeed rapidly rising toward the red line. Now what to do?

@TypeIA -- one more key point-- re your comment " (3) is the point of minimum drag." -- actually it is the point of minimum drag coefficient.  It can be shown that for steady-state flight, drag force is minimized at the point of max L/D or max Cl/Cd.  Likewise, point #5 is the point of max lift coefficent, but the only way to say that this also corresponds to the point of max lift force, is to constrain the airspeed to somehow magically stay constant, as other parameters (such as A-o-A) vary.

@UXNoob4eva -- point 5 is Cl max. The blue line can't be described as the curve of "Total Drag". Rather it gives the lift coefficient and the drag coefficient of any point on the line. The left-right or horizontal position of any point on the blue line describes the drag coefficient at that point. But that is not the same as the Total Drag at that point. See my other comment immediately above for more.

To be more exact, the above figure is taken from a question catalog for the AustroControl PPL theory exam. The catalog is based on the de-facto standard German PPL textbook
Advanced PPL-Guide 2a - Aerodynamik Flugzeuge  by Aircademy Ldt.
The book itself including the figure is under copyright, so it is best not to post a scan here. The original annotations for the numbers of that figure in the book (Chapter 2, "Strömung am Tragflügel", Fig. 19) are:

number
original annotation (DE)
translated annotation (EN)

1
Rückenflug
inverted flight

2
$c_a = 0$, Sturzflug
nose dive

3
$c_w = min$

4
$\frac{c_a}{c_w} = max $

5
$c_a = max$, Langsamflug
slow flight

6
überzogener Flugzustand
stall

Did you provide the English translations of the annotations or are they in the original? "Nose dive" is a surprising / harsh description of zero load factor. (For that matter I'm curious about "Sturzflug": I speak German, but my flight training and experience is USA-based so I'm not sure if this really is the standard term for this scenario.) Nice answer though, +1 for identifying the source material.

The original annotations are in German („Sturzflug“, etc.). English translations my own (cross-checked at dict.cc).

@TypeIA -- echoes of my comments to the actual question-- zero lift / zero load factor/ zero G's can certainly be demonstrated or experienced in conditions other than a vertical dive, but a vertical dive ("nose-dive") is the only situation where  zero lift / zero load factor/ zero G's (as registered on the G-meter which only detects the "up-and-down" component in the aircraft's normal reference frame)  can be sustained in a steady-state situation. It's a bit much to expect study materials for a PPL exam to be cognizant of the differences between steady-state and transient conditions.

The graph seems to make more sense for a non-symmetrical airfoil.
1 is clearly a negative angle of attack
2 is no lift but slightly more drag than 3 ... hmm
3 has lift but a little less drag.  Non symmetrical airfoils can generate lift at 0 Angle of Attack, so 2 probably needs to be at a slightly negative AoA for zero lift.  This accounts for the higher drag.
4 is best lift to drag ratio
5 is a higher Coefficient of Lift but proportionally higher drag, which means higher AoA to compensate for slower flight (or higher wing loading, such as in a turn).  Point 5 would be at what is known as "stall speed" (really stall AoA).
6 yes, this is stalled

Re "2 is no lift but slightly more drag than 3 ... hmm"-- yes, as you noted, this would be for a non-symmetrical (cambered) airfoil.  To make zero lift, despite the cambered shape, the airfoil must be plowing along nose-down at a negative angle-of-attack-- more draggy than an A-o-A closer to zero, which would generate some mild amount of lift.  PS I'm not sure what is the significance of the "Lilienthal" reference, but all his wings were highly cambered!  (Very thin, with strongly concave under-surfaces.)


text:
Why does the climb angle of 737-800 peak shortly after take-off before reducing again?
I am doing a report on 737-800 and I have questions about its climb angle performance;
I am plotting a climb angle VS altitude graph for the aircraft, why does the climb angle take off at an angle and then increase to a maximum value and then decrease again as it ascends?
Does anyone know the climb angle that the aircraft takes off at? and the maximum climb angle of the aircraft?
If could provide a factual link would be much appreciated.

@user65712 could you post a picture of the climb graph? I have some ideas for what could explain the behavior you describe, but I'd need to see the original data.


text:
Where is the official definition of pressure altitude?
I know that pressure altitude for a certain pressure is defined to be the altitude in the standard atmosphere which has that pressure.
But I can't find whatever standard, regulation, or glossary officially defines it to be so. I tried to look in the AIM, the P/CG, a search of the FARs for "pressure altitude" wading through a ton of results, and what searchable partial copies of the US Standard Atmosphere of 1976 I could find. I don't have access to a copy of the ISO standard that defines the International Standard Atmosphere.
Does anyone know of an FAA, ICAO, or ISO document providing this definition?

There's an FAA document that contains a definition at https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/pilot_handbook/media/PHAK%20-%20Chapter%2003.pdf.

An unanswered part of this question asks what document defines the International Standard Atmosphere. The answers below do define pressure altitude in terms of it, but the official definition of the international standard atmosphere is located in ICAO Doc 7488.

Pressure Altitude
As defined by FAA in the Pilot's Handbook of Aeronautical Knowledge:

Pressure altitude is the height above a standard datum plane (SDP),
  which is a theoretical level where the weight of the atmosphere is
  29.92 "Hg (1,013.2 mb) as measured by a  barometer. An altimeter is essentially a sensitive barometer calibrated to indicate altitude in
  the standard atmosphere. If the altimeter is set for 29.92 "Hg SDP,
  the altitude indicated is the pressure altitude. As atmospheric
  pressure changes, the  SDP may be below, at, or above sea level.
  Pressure altitude is important as a basis for determining airplane
  performance, as well as for assigning flight levels to airplanes
  operating at or above 18,000 feet.
The pressure altitude can be determined by either of two  methods:
  1. Setting the barometric scale of the altimeter to 29.92 and reading the indicated altitude.
  2. Applying a correction factor to the indicated altitude according to the reported altimeter setting.

More info from the FAA here.
From Docket No. 18334, 54 FR 34304, Aug. 18, 1989, as amended by Amdt. 91-314, 75 FR 30193, May 28, 2010

§91.217 Data correspondence between automatically reported pressure altitude data and the pilot's altitude reference.
(a) No person may operate any automatic pressure altitude reporting equipment associated with a radar beacon transponder—
(1) When deactivation of that equipment is directed by ATC;
(2) Unless, as installed, that equipment was tested and calibrated to transmit altitude data corresponding within 125 feet (on a 95 percent probability basis) of the indicated or calibrated datum of the altimeter normally used to maintain flight altitude, with that altimeter referenced to 29.92 inches of mercury for altitudes from sea level to the maximum operating altitude of the aircraft; or
(3) Unless the altimeters and digitizers in that equipment meet the standards of TSO-C10b and TSO-C88, respectively.
(b) No person may operate any automatic pressure altitude reporting equipment associated with a radar beacon transponder or with ADS-B Out equipment unless the pressure altitude reported for ADS-B Out and Mode C/S is derived from the same source for aircraft equipped with both a transponder and ADS-B Out.

ICAO has described how they calculate Pressure Altitude

Source
In relation to QNE:

When the ISA mean sea level standard pressure of 1013.2 hPa is set on
  an aircraft altimeter subscale, the height so indicated upon landing
  at an airfield is known as the QNE reading. More widely, this is also
  the PRESSURE ALTITUDE, which is alternatively defined as the height of
  any level in the international standard atmosphere (ISA-see above),
  above the level corresponding to a pressure of 1013.2 hPa.

Here is a free tool to calculate standard atmospheric conditions (ISA) at a given geometric or pressure altitude and ambient temperature.

however comprehensive this answer is, I don't think any of the sources cited are authoritative. Even the PHAK is not regulatory.

The notion of pressure-altitude, used in aviation and displayed on barometric altimeters, is set by ICAO. It is derived from a model elaborated by WMO and ICAO, two agencies of the United Nations, building on previous work done in the US and in Europe.
ICAO defines pressure-altitude in distinct documents:

ICAO Annex 2 of the Chicago Convention (Rules of the Air)
ICAO Annex 8 of the Chicago Convention (Airworthiness of Aircraft)
ICAO Doc 4788 (Manual of the ICAO Standard Atmosphere)

In addition, ICAO atmosphere model has been published as ISO 2533 standard, subtitled "identical with the ICAO and WMO Standard Atmospheres from -2 to 32 km", hence compatible with ICAO model. ISO is interlocked with ICAO for updates.
Annex 2 - Rules of the Air
Annex 2 introduces the notion of pressure-altitude and where it is defined:

Pressure-altitude. An atmospheric pressure expressed in terms of altitude which corresponds to that pressure in the Standard Atmosphere. As defined in Annex 8.

This is a pressure measurement converted into an equivalent height above mean see level, according to the model used by ICAO.
Annex 8 - Airworthiness of Aircraft
This document provides a description of what is the standard atmosphere to be used when designing and certifying aircraft performances, e.g. for takeoff and landing at high altitude airports.
Some constant and gradients are provided:

The model and various formulas are fully detailed in the last document.
Doc 7488 - Manual of the ICAO Standard Atmosphere
Details and precomputed tables are provided in ICAO Doc 7488 for various parameters. In particular:

Assuming a linear variation of the temperature with geopotential altitude, the simultaneous solution of the hydrostatic equation (equation (1)) and the perfect gas law (equation (2)) yields the following expression for pressure:
[...]
$$p = p_b \; exp \bigg [ \frac {g_0} {RT}(H-H_b) \bigg ] \quad (13)$$

From in this formula we can derive the relationship pressure-to-altitude. For altitudes below 11km (p in hPa, h in m):
$$h = 44301.59796 \times (1 - (\frac {p} {1013.25})^{0.190284})$$
or equivalent (formula from ISO 2533):
$$h = \frac {3.731444 - p^{0.1902631}} {0.841728  \times 10^{-4}}$$
This gives this curve:

The formulas constants change at 11, 20 and 32 km, following the different temperature gradients in the different atmosphere layers.
Coefficient and exponent values reflect the different factors taken into account to determine the Standard Atmosphere model, including sea level pressure, temperature, density and gravity gradients (see this answer from @fab for details). However this model is simplified, e.g. air is always dry.

This is the best definition of pressure altitude. This should be the official answer to the question

If you want a more detailed equation to compute pressure altitude you can check this post: https://aviation.stackexchange.com/questions/82377/how-can-i-calculate-ambient-air-pressure-from-flight-data#95460

@fab: Excellent addition, it shows how the different parameters are merged into the final formula. So everybody interested should also read your answer.


text:
How can the Moment Coefficient with elevator deflection be obtained?
For the range of linear lift, the canard moment coefficient about the aerodynamic center is:
$C_{m_c}=C_{m_c0} +C_{m_c,\delta_e}\delta_e $
Where

$C_{m_c0}$ is the pitching moment coefficient without any control surface deflection, and its value can be found by using the airfoil data.
$C_{m_c,\delta_e}$ is the change in moment coefficient with respect to elevator deflection.
$\delta_e$ is the elevator deflection angle.

Is there a way to find the value of the second term $C_{m_c,\delta_e}$?

As usual in those cases the USAF STABILITY AND CONTROL DATCOM is the way to go.
In particular, section 6.1.2. at page 1945 of the .pdf contains the methods that you need. At page 1949 a very simple method is presented suitable for plain flap/elevator:

Pitching-moment increments for plain trailing-edge flaps are presented in Figure 6.1.2.1-35b. These increments are given about the quarter-chord, based on the square of the airfoil chord c², as a function of flap deflection and the ratio of flap chord to airfoil chord cf/c. This figure is limited to the linear-lift range and subcritical Mach numbers.

Thank you, Sophit  much better now.

@DeltaLima: you're welcome


text:
Is there any website which is offering free real time tracking data of flights?
Our customer always demand, and I think there is huge market of, live tracking data. We need to ease either our customers or their relatives that they have reached their destination safely. So as I was moving around, I found out that there are too many live data APIs, but I want to integrate the free live API in order to check first. Can any one give me options that provide live APIs? Also if I need to get ADS-B what can I do and how does it work?

I've edited your question to make it a bit more readable, but I am not fully sure of what you mean in a couple of sentences. At first impression, this question seems a bit broad/opinion-based for this website, be aware that we are not a discussion forum. Please have a look on how to ask

As this need is quite common today, I guess this is now a programming question (finding and using related API) rather than an aviation one. Random sample, also on this site. As for how it's used in the code, here is the interface reference for Flightstats.

There are numerous web sites that provide flight tracking data, including radar and ADS-B. Most are free for personal use, but AFAIK all of them charge for commercial use. This makes sense since collecting and aggregating that data takes a lot of time, effort and equipment, and they have to recover those costs somehow. Still, using an existing service should be a lot less expensive than trying to set up your own redundant operation just for your own needs.
The sole exception is if you're trying to track ADS-B flights in the vicinity of your own facility, in which case you can easily set up a single ADS-B receiver there. But most folks want a (much) bigger picture, including non-ADS-B traffic, so please clarify the scope of your needs.

On the ADS-B side of things, if your looking for a purely personal system at receiving ADS-B, a Raspberry Pi 3b, a USB dongle (RTL2382 version), a antenna and some coax - install Dump1090 onto the Pi and your in action. Not this simple but a basic overview.
If you are looking for a turn-key 'plug and play' solution you might want to check out VariFlight (VariFlight ADS-B) , they often provide free ADS-B systems with everything you need, you just plug it in and go. The only requirement is you have the ADS-B receiver connected to the internet to feed them the data you are receiving.


text:
Why is it called "Dry Thrust"?
When reading the specifications of a jet aircraft anywhere on the Internet, it usually states that its engine produces X pounds of "dry thrust".
Why is it called "Dry" Thrust? As opposed to what?

Some information about afterburners: http://aviation.stackexchange.com/questions/17286/what-is-an-afterburner-and-how-long-can-a-jet-fly-on-afterburner/17287#17287

Dry thrust usually means the non-augumented thrust i.e. thrust without the use of afterburners or liquid injection. The maximum thrust produced by jet engines w/o afterburner is sometimes called military thrust.
The thrust of a jet eingine can be increased by using methods like water(+methonol) injection (mostly in older turbojet engines) or by using afterburners (reheat). In such cases, the (higher) thrust produced is called wet thrust.

Dry thrust refers to an afterburning engine running without afterburner.  The figures you mention usually refer to the maximum thrust without afterburner.

One interesting case of actually wet thrust, as opposed to afterburners, is the Harrier Jump Jet, which can inject water into its engines for up to 90 seconds to increase performance for vertical takeoff and landing without melting its turbine blades. The use of water in jet engines is interesting because it not only cools the turbine, but simultaneously adds thrust through the added mass and steam expansion.


text:
Centrifugal force created by a 1045 propeller
I currently am creating a project in which I'm using a 1045 propeller with a BLDC motor that spins at ~16,000RPM. It is powering a single motor "rocket" that basically is configured to carry a payload to ~380m. I've done a lot of testing with the rocket and found that it's reaching all the altitudes and speeds correctly. However, the rocket spins quite a lot in the roll axis. The payload that I'm carrying is actually meant to benefit from the centrifuge that the rocket is acting like but I would like to calculate the exact centrifugal force created by the propeller.
$$F = \frac{m v^2}{r}$$
I used this formula for calculating the centrifugal force created by the propeller. Since I know the angular velocity (16,000RPM) of the propeller, I calculated the tangential velocity to be 207.76m/s.
Mass of propeller - 28g
Velocity =  217.8m/s
Radius = (25.4/2) = 12.4
I got, F = (0.028) x (207.76)² /  0.124
Which gave me, F = ~9747N. That's 35,498g of acceleration.
I personally think something is wrong with that, isn't it excessive? Or am I just mistaken?

When the propeller spins, it makes the rocket spin the other way around as per Newton's third law. To calculate the two mutual rotating speeds you need to apply Newton's second law and that's why I think this question is better served on physics Stackexchange

Problems I found:

You are calculating the centrifugal force of the propeller, not the payload (unless the payload is spinning with the propeller - or perhaps if payload is the propeller itself).

Radius is incorrect. When calculating centrifugal force, "radius" is not simply the shaft-centre to blade-tip distance - it is the distance between shaft-centre and the CG of the prop blade.

Mass is incorrect. It should be 14g - the mass of one blade, not the entire prop. (And regardless, it is supposed to be payload mass, not prop mass).

Suggestions:
In the question, it appears that the rocket is spinning due to the propeller torque (like a helicopter without tail rotor). You first need to determine the rotation speed of the vehicle - excluding the prop.
Next, you need to determine the CG of the payload, and that of the rocket with the payload installed. The distance between the two CG's is the radius. From here, you can determine the tangential velocity of the payload. Finally, measure the payload mass and from there you can calculate the centrifugal force.
However, this is only helpful if you assume a point-sized payload. For accuracy, I would instead recommend that you directly calculate the acceleration at the part of the payload you're concerned with (unless the payload size is small compared to the radius).

Determining vehicle rotation speed
For obvious reasons, this is no easy task; so here are some solutions.

Determine Experimentally: Attach the vehicle to a pivot on its rolling axis, and select the normal flight power. To prevent vehicle breakdown, consider encasing the vehicle all around with a structure that may arrest it. The steady state vehicle RPM in this case will almost certainly be higher than what it would be in flight. If available, you can use a powerful fan blowing onto the vehicle from above - to simulate the real flight.

Make a simple RPM sensor: Install an LDR (Light Dependent Resistor) to a side of your rocket, figure out a way to accurately time its output pulses, and launch the rocket in morning/evening on a sunny day. If all goes well, the number of pulses in a given time should give you the RPM.

The rocket is spinning due to the propeller being at the top. I wanted to calculate the centrifugal force that the payload is being subjected to. And yes, the rocket spins just like a helicopter with a missing tail rotor. Determining the rotation speed would be quite complicated for me, so I thought that I could calculate the rotation speed.

@AadirajAnil I understand how complicated it would be to determine the rotation speed, but I can assure you that prop RPM would be quite different from the vehicle RPM. Also, I didn't get why you used prop radius and prop mass in your calculation, where you should have instead used payload radial distance as "r" and payload mass as "m". (Also, I have an idea to calculate the vehicle rotation speed - it's kind of a jugaad but it might give a good approximation, let me know if I can help.)

I used prop radius since I'm new in the area and was confused. It would be great if you could help with the vehicle RPM, even if it's a jugaad!

@AadirajAnil Alright, give me some time, I will add that to my answer :-)

@AdityaSharma and Aadiraj: to get the rotating speed of the rocket's body, CG is not enough, you need the moments of inertia of the propeller and the rocket's body. Afterward you can apply Newton's second law to estimate the two rotating speeds. Or you just do some testing but I don't know if it's easier (for sure more funny).

@sophit lol who's gonna determine the moment of inertia and all that stuff when you can jerry rig an LDR to that rocket ;-)

I suppose the rocket is a simple cylinder. And the moment of inertia of a body can be simply determined making the body swing and measuring the period of the oscillation.

@sophit You are absolutely right; I was just joking that why get into the math when when an LDR gets the job done :-) good day!

@AdityaSharma yes sure, testing is much funnier than maths

A similar issue was observed with a solid fuel model rocket.  The good news is that spin can be stopped and even controlled to a specific rate with fins.
The spinning forces here are not unlike those found in single prop aircraft and helicopters.  What makes it particularly interesting is that, with the high RPM of the prop, and less rotational drag resistance, the rocket spin rate may become quite high.
First, control it, quantify it, then adjust it by fine-tuning how much spin resistance you want.
A limiting factor here is, as spin rate increases, prop thrust decreases because relative to the free stream of incoming air, prop rpm slows down.  Even though 16,000 rpm is quite high, you may have noticed a decrease in the rockets forward speed as it spins up.
With fins, rotational motion of the airframe or body they are attached to changes their relative wind, giving them anti-spin lifting force qualities.

The spin will reach steady state rotational velocity when prop/engine torque is balanced by fin/body resistance torque along their respective radii.

This is bound to happen quickly as torque is great, and mass and radius are low. Rotational V zero to Rotational steady state may be on the order of a few seconds.
As in the early days of full scale rocketry, alternating stripes can be added on so photography (back then frames per second) could be used to determine spin rate.